【问题标题】:Jenkins groovy pipeline search and compareJenkins groovy 管道搜索和比较
【发布时间】:2018-06-26 15:33:31
【问题描述】:

我需要一个解决方案来解决以下问题 - 在ver_list 中搜索值变量BRANCHVERSION,比较它们并使用合适的。

BRANCHVERSION 的格式类似于"6.200.01"

​ver_list 看起来像这样

[Pipeline] echo
[6.000.02, 6.000.10, 6.000.12, 6.000.15, 6.000.20, 6.000.25, 6.000.30, 6.100.00, 6.100.01, 6.100.10, 6.100.20, 6.100.25, 6.100.30, 6.200.00, 6.300.00]

如果BRANCHVERSION 不等于ver_list 中的任何值,则必须从按降序排列的列表中选取下一个最接近的值(如果BRANCHVERSION = 6.200.01,则必须为6.200.00已接) 如果BRANCHVERSION 等于ver_list 中的任何值,则选取该值。

我的管道描述如下:

if (ENVIRONMENT == "") {
   error("You should choose at least one environment")
}
node {   
   deleteDir()
    checkout([$class: 'SubversionSCM', locations: [[credentialsId: 'XXXXXXX',local: './env_prop_files', remote:'http://FQDN/repos/src/cm/env_prop_files']]])
    checkout([$class: 'SubversionSCM', locations: [[credentialsId: 'XXXXXXX',local: './SoapUITestSuites', remote:'http://FQDN/repos/src/tools/SoapUI/branches/']]])
    def  FILES_LIST = sh (script: "ls   './SoapUITestSuites'", returnStdout:true).trim()
    def GS_LIST = sh (script: "ls 'path_to_file'", returnStdout: true).trim()
echo "GS_Version : ${GS_LIST}"
echo "FILES_LIST : ${FILES_LIST}"
//PARSING TestSuite Version
   def ver_list = []
     for(String ele : FILES_LIST.split("\\r?\\n")){ 
     ver = ele.split("_")
     println ">>>${ver[1]}<<<"     
     ver_list.add(ver[1])
}
println(ver_list)
def VERSION = ver_list
//PARSING GS Version
   def gs_list = []
    for(String elegs : GS_LIST.split("\\r?\\n")){ 
    println ">>>${elegs}<<<"     
    gs_list.add(elegs)
}  
println(gs_list) 
properties = readProperties file:"./env_prop_files/${ENVIRONMENT}.properties"
BRANCHVERSION = properties.SVN_LOCATION_VERSION
}

输出

No changes for http://FQDN/repos/src/tools/SoapUI/branches since the previous 
build
[Pipeline] sh
[ClientRunSoapUITest] Running shell script
+ ls ./SoapUITestSuites
[Pipeline] sh
[ClientRunSoapUITest] Running shell script
+ ls /pkg/flexprod/oracle/flexprodfiles/archive/goldstand
[Pipeline] echo
GS_Version : 5.900.00
6.000.00
6.000.01
6.000.02
6.000.20
6.000.30
6.100.00
6.100.01
6.200.00
[Pipeline] echo
FILES_LIST : rel_6.000.02
rel_6.000.10
rel_6.000.12
rel_6.000.15
rel_6.000.20
rel_6.000.25
rel_6.000.30
rel_6.100.00
rel_6.100.01
rel_6.100.10
rel_6.100.20
rel_6.100.25
rel_6.100.30
rel_6.200.00
rel_6.200.01
rel_6.300.00
[Pipeline] echo
>>>6.000.02<<<
[Pipeline] echo
>>>6.000.10<<<
[Pipeline] echo
>>>6.000.12<<<
[Pipeline] echo
>>>6.000.15<<<
[Pipeline] echo
>>>6.000.20<<<
[Pipeline] echo
>>>6.000.25<<<
[Pipeline] echo
>>>6.000.30<<<
[Pipeline] echo
>>>6.100.00<<<
[Pipeline] echo
>>>6.100.01<<<
[Pipeline] echo
>>>6.100.10<<<
[Pipeline] echo
>>>6.100.20<<<
[Pipeline] echo
>>>6.100.25<<<
[Pipeline] echo
>>>6.100.30<<<
[Pipeline] echo
>>>6.200.00<<<
[Pipeline] echo
>>>6.200.01<<<
[Pipeline] echo
>>>6.300.00<<<
[Pipeline] echo
[6.000.02, 6.000.10, 6.000.12, 6.000.15, 6.000.20, 6.000.25, 6.000.30, 6.100.00, 6.100.01, 6.100.10, 6.100.20, 6.100.25, 6.100.30, 6.200.00, 6.200.01, 6.300.00]
[Pipeline] echo
>>>5.900.00<<<
[Pipeline] echo
>>>6.000.00<<<
[Pipeline] echo
>>>6.000.01<<<
[Pipeline] echo
>>>6.000.02<<<
[Pipeline] echo
>>>6.000.20<<<
[Pipeline] echo
>>>6.000.30<<<
[Pipeline] echo
>>>6.100.00<<<
[Pipeline] echo
>>>6.100.01<<<
[Pipeline] echo
>>>6.200.00<<<
[Pipeline] echo
[5.900.00, 6.000.00, 6.000.01, 6.000.02, 6.000.20, 6.000.30, 6.100.00, 6.100.01, 6.200.00]
[Pipeline] readProperties
[Pipeline] }
[Pipeline] // node
[Pipeline] echo
"6.200.01"
[Pipeline] lock

【问题讨论】:

    标签: shell jenkins groovy jenkins-pipeline jenkins-groovy


    【解决方案1】:

    首先,这些是字符串,所以你应该c将它们挂起来。比较它们要容易得多:6.200.21 变为 6200,21(或 6200.21,具体取决于您的本地化)

    将它们保存在一个列表中,这样处理它们会容易得多 剩下的就是简单的 Groovy List 操作

    BRANCH_VERSION = ver_list.contains(BRANCH_VERSION) ? BRANCH_VERSION : ver_list.max()
    

    当然,如果您需要将其作为字符串,则使用 DecimalFormat 对象将其转换回来并用点替换逗号。

    【讨论】:

    • OP 想要下一个最接近指定的数字,因此您必须添加一个函数来获取下一个最接近的值,如下逻辑 *.com/questions/13318733/…
    • 对。我想知道 lambda 函数等是否有更时髦的方法,比如创建一个增量列表:def deltas = ver_list.collect { (it - BRANCH_VERSION ).abs()}。之后,您可以使用def closest_release = ver_list.filter { (it - BRANCH_VERSION).abs() == deltas.min()}.last() 也就是与 BRANCH_VERSION 距离最小的最后一个(Descending)项目获得下一个关闭项目。 ...最后,表达式简化为:BRANCH_VERSION = ver_list.contains(BRANCH_VERSION) ? BRANCH_VERSION : closest_release
    • 是的,有可能...Groovy 使用与 Java 的 Lamda 非常相似的闭包
    • 我是如何解决这个问题的: def versiongs = sh (script: "ls '/pkg/flexprod/oracle/flexprodfiles/archive/goldstand' | LC_ALL=C awk '\$0